1
$\begingroup$

I have been trying to prove variational theorem in quantum mechanics for a couple of days but I can't understand the logic behind certain steps. Here is what I have so far: \begin{equation} E=\frac{\langle \phi ^*|H|\psi\rangle}{\langle \phi ^*|\phi\rangle} \end{equation} consider a wavefunction close to the ground state, \begin{equation} \psi=\psi_0+\delta \psi \end{equation} \begin{equation} E=\frac{\langle\psi ^*_0+\delta \psi^* |H|\psi _0+\delta \psi\rangle}{\langle \psi ^* _0 +\delta \psi ^*|\psi _0+\delta \psi \rangle} \end{equation} (here is where I begin to struggle ... I don't understand the mathematical step). \begin{equation} =\frac{\langle \psi _0^*|H|\psi _0\rangle+2\langle\delta\psi^*|H|\psi _0\rangle+\langle\delta\psi^*|H|\delta\psi\rangle}{\langle\psi^*_0|\psi_0\rangle+2\langle\delta \psi^*|\psi_0\rangle+\langle\delta\psi^*|\delta \psi\rangle} \end{equation} \begin{equation} =\frac{E_0\langle \psi _0^*|\psi _0\rangle+2E_0\langle\delta\psi^*|\psi _0\rangle+\langle\delta\psi^*|H|\delta\psi\rangle}{\langle\psi^*_0|\psi_0\rangle+2\langle\delta \psi^*|\psi_0\rangle+\langle\delta\psi^*|\delta \psi\rangle} \end{equation} \begin{equation} =E_0 +O((\delta\psi)^2) \end{equation} I understand that the change in the wavefunction leads to an energy term that is second order, yet how can we tell that the best wavefunction is the one that minimises this energy? Is it because we ignore the $\delta ^2$ term?

There is another alternative proof here which I also can not follow. http://www.nyu.edu/classes/tuckerman/quant.mech/lectures/lecture_3/node1.html

$\endgroup$
  • $\begingroup$ remember that the variational method is used in near-unperturbed type of approximations. Where the overall state is very close to the state of the free / un-perturbed system $\endgroup$ – Nikos M. Oct 22 '14 at 17:08
2
$\begingroup$

I don't find this proof a good one, since the notation is messy and not very clear (not to say wrong). One proof can be given in a similar way to the one you posted in the link. Suppose the spectrum of $H$ is discrete and the set of eigenstates $\{|\phi_n\rangle\}$ constitutes an orthonormal basis with eigenvalues $E_n$, such that $E_0\leq E_1\leq E_2\leq\dots$. So for any normalized state $|\psi\rangle$, we can expand it in this base: $$|\psi\rangle=\sum_nc_n|\phi_n\rangle$$ Then we have \begin{align}\langle\psi|H|\psi\rangle&=\left(\sum_mc_m^*\langle\phi_m|\right)H\left(\sum_nc_n|\phi_n\rangle\right)\\ &=\sum_{m,n}c_m^*c_n\langle\phi_m|H|\phi_n\rangle\\ &=\sum_{m,n}c_m^*c_n\langle\phi_m|E_n|\phi_n\rangle\\ &=\sum_{m,n}c_m^*c_nE_n\langle\phi_m|\phi_n\rangle \\ &=\sum_{m,n}c_m^*c_nE_n\delta_{mn}\\ &=\sum_n|c_n|^2E_n\\ &\geq \sum_n|c_n|^2E_0=E_0, \end{align} since, $\sum_n|c_n|^2=1$ and $E_n\geq E_0$, where $E_0$ is the lowest eigenstate of $H$. It can be proved that this theorem also holds in the case that there is a lowest eigenvalue $E_0<\sigma_{ess}(H)$ in the spectrum of $H$, even though the spectrum is not made only of eigenvalues.

$\endgroup$
0
$\begingroup$

Let's run through the variational principle very quickly. The idea is that an arbitrary state $\psi$ can be decomposed into a sum of orthogonal energy eigenstates:

$\left|\psi\right> = \sum c_n \left|\psi_n\right>$ where $\sum |c_n|^2 = 1$ and $H\left|\psi_n\right> = E_n \left|\psi_n\right>$

Then the expectation of the energy $\left<E\right>$ is:

$ \left<E\right> = \left<\psi\right|H\left|\psi\right> = \left<\psi\right|\sum c_n E_n\left|\psi_n\right> = \sum c_nc_m^* E_n\left<\psi_m|\psi_n\right> = \sum |c_n|^2 E_n$

The rightmost expression is just a weighted average of the energies of each energy eigenstate $\left|\psi_n\right>$. If there is a minimum energy $E_0$, then $\left<E\right>$ is clearly minimized when all the coefficients are zero except for $c_0=1$, so that $\left|\psi\right> = \left|\psi_0\right>$.


I think your notation would be clearer if you represented states with coefficients multiplying normalized states, eg $\left|\psi\right> = a\left|\psi_0\right> + b\left|\delta\psi\right>$. In your current notation none of the states are normalized which makes everything harder. Also you shouldn't write the complex conjugate $^*$ within the left state vectors, but they do apply to the coefficients: $\left<\psi\right| = a^*\left<\psi_0\right| + b^*\left<\delta\psi\right|$.

As for the step your are struggling with, $\left|\delta\psi\right> = \sum_{n>0} c_n \left|\psi_n\right> $ represents all the components of the state $\left|\psi\right>$ that are not the ground state $\left|\psi_0\right>$. This means it is orthogonal to the ground state which will simplify your expressions. This also means that $E_\delta = \left<\delta\psi\right|H\left|\delta\psi\right>$ is necessarily greater than $E_0$. The expectation value of the energy is then

$\left<E\right> = a^2E_0 + b^2E_\delta$

which is minimized for $b=0$. This just tells you that the minimum energy state is (by definition really) the ground state. So you can determine the approximate ground state by twiddling with a test state until you've minimized its energy. There are numerical algorithms to do the twiddling and minimizing for you.

$\endgroup$
  • $\begingroup$ If we always try to minimise the energy how come we don't always get the ground state ? $\endgroup$ – user58536 Oct 22 '14 at 21:34
  • $\begingroup$ Because when you try to minimize, you are always restricted to specific functions that you vary with respect to some parameter. So, even if you minimize the energy with respect to this parameter, you generally aren't able to find the correct eigenfunction. E.g. Suppose you have the harmonic oscilator. If you chose $\psi_a(x)=C\dfrac{1}{1+ax^2}$, when you minimize $E=\dfrac{\langle \psi|H|\psi\rangle}{\langle \psi|\psi \rangle}$ you don't get the right eigenfunction, although you will have an specific $a=a_0$ that minimizes it for all $\psi_a(x)$. $\endgroup$ – Mateus Sampaio Oct 22 '14 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy